Prove that you can choose 2k numbers from the set {1, 2, 3, …, 3k−1} in such a way that the chosen set contains no averages of any two of its elements.

Let me show how to find 2k − 1 such numbers. We can pick all the numbers that have 0 or 1 in their ternary representation. Let me prove that this set doesn’t contain averages. Summing two such numbers doesn’t involve carry, and the sum will contain a 1 in each place where the digits differ. On the other hand, a double of any number in this set doesn’t contain ones.

This solution is pretty, but it is not good enough: we need one more number. We can add the number 3k−1. I will leave it to the reader to prove that the largest number in the group whose binary representation consists only of twos can be added without any harm.

There are other ways to solve this problem. It is useful to notice that multiplying a no-average set by a constant or adding a constant to it, doesn’t change the no-average property.

If we were allowed to use 0, then the problem would have been solved. As zero doesn’t belong to the initial range, we can add zero and shift everything by 1. The resulting sequence is sequence A3278. This sequence is the lexicographically first non-averaging sequence.

Another solution was suggested by devjoe in my livejournal mirror blog site. If we multiply our non-averaging set (the one that doesn’t have twos in their ternary representation) by 2, we get a set of all numbers that do not contain ones in their ternary representation. By linearity, such a set doesn’t contain averages either. We can add 1 to this set.